\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,1           0,2            0,1      0,1 
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\\ V_{H_2}=0,1.22,4=2,24l\\ m_{\text{dd}}=6,5+200-\left(0,1.2\right)=206,3g\)  
bài 2 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\) 
          0,2             0,4       0,2              0,2 
\(m_{HCl}=0,4.36,5=14,6g\\ V_{H_2}=0,2.22,4=4,48l\\ m\text{dd}=4,8+200-0,4=204,4g\\ C\%=\dfrac{0,2.136}{204,4}.100\%=13,3\%\)

a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Ta có: \(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

Theo PT: \(n_{Mg}=n_{H_2}=0,03\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,03.24=0,72\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,72}{1,74}.100\%\approx41,38\%\\\%m_{AlCl_3}\approx58,62\%\end{matrix}\right.\)

b, Theo PT: \(n_{HCl}=2n_{H_2}=0,06\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,06.36,5=2,19\left(g\right)\)

\(\Rightarrow C\%_{ddHCl}=\dfrac{2,19}{500}.100\%=0,438\%\)

Bạn tham khảo nhé!

a, nH2 = 0,03 ( mol )

=> nMg = nH2 = 0,03 ( mol )

=> mMg = 0,72 g

=> %Mg \(\approx\) 41,38 % .

=> % Al \(\approx\) 58,62 % .

b, Có : nH2 = 0,03 mol

=> nHCl = nHCl_{từ Al2O3} + nHCl_{từ Mg} = 0,06 + 0,06 = 0,12 ( mol )

=> mHCl = 4,38 ( g )

Lại có : mdd = mhh + mddHCl = 501,74 ( g )

=> \(C\%=\dfrac{m_{HCl}}{m_{dd}}.100\%\approx0,87\%\)

( chắc đoạn trên là Al2O3 :vvvv )

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(0.2...................0.2..........0.2\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{FeCl_2}=0.2\cdot127=25.4\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=11.2+200-0.2\cdot2=210.8\left(g\right)\)

\(C\%_{FeCl_2}=\dfrac{25.4}{210.8}\cdot100\%=12.05\%\)

Giúp với mai mình thi rồi!

a/ Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂

n_H2 = \(\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PTHH: n_H2 = n_Fe = 0,15 (mol) \(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

\(\Rightarrow m_{Cu}=11-8,4=2,6\left(g\right)\)

\(\Rightarrow\%m_{Fe}=\dfrac{8,4}{11}.100\%\approx76,4\%\)

\(\Rightarrow\%m_{Cu}=100-76,4\approx23,6\%\)

b/ Theo PTHH ta có: n_HCl = 2n_Fe = 2.0,15 = 0,3 (mol)

\(\Rightarrow V_{ddHCl}=\dfrac{0,3}{2}=0,15\left(M\right)\)

c/ m_HCl = 36,5 . 0,3 = 10,95(g)

\(\Rightarrow C\%_{HCl}=\dfrac{m_{HCl}}{m_{ddHCl}}.100\%=\dfrac{10,95}{200}.100\%=5,475\%\)

a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,15.56}{11}.100\%\approx76,36\%\\\%m_{Cu}\approx23,64\%\end{matrix}\right.\)

b, Theo PT: \(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{0,3}{2}=0,15\left(l\right)\)

c, \(C\%_{HCl}=\dfrac{0,3.36,5}{200}.100\%=5,475\%\)

\(a)2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ b)n_{H_2}=\dfrac{5,6}{22,4}=0,25mol\\ n_{Al}=a;n_{Fe}=b\\ \left\{{}\begin{matrix}3a+b=0,25\\27a+56b=8,3\end{matrix}\right.\\ a=\dfrac{19}{470};b=\dfrac{121}{940}\\ \%m_{Al}=\dfrac{\dfrac{19}{470}\cdot27}{8,3}\cdot100=13,15\%\\ \%m_{Fe}=100-13,15=86,85\%\\ c)n_{HCl}=3\cdot\dfrac{19}{470}+2\cdot\dfrac{121}{940}=\dfrac{89}{235}mol\\ m_{ddHCl=}=\dfrac{\dfrac{89}{235}\cdot36,5}{7,3}\cdot100=189g\\ d)n_{AlCl_3}=n_{Al}=\dfrac{19}{470}mol\\ n_{Fe}=n_{FeCl_2}=\dfrac{121}{940}mol\)

\(m_{dd}=8,3+189-0,25.2=196,8g\\ C_{\%AlCl_3}=\dfrac{\dfrac{19}{470}\cdot133,8}{196,8}\cdot100=2,8\%\\ C_{\%FeCl_2}=\dfrac{\dfrac{121}{940}127}{196,8}\cdot100=8,3\%\)

Fe + 2HCl -> FeCl2 + H2

1         2            1         1

0,5      1            0,5      0,5 

n_Fe= m/M = 26.3 / 56 = 0,5 mol

a) VH= n . 22,4 = 0,5 . 22,4 = 11,2 lít

b) m_{ct HCl}= n . M = 0,5 . 36,5 = 18,25g

C% = ( m_{ct . 100 )/ mdd} =18,25 . 100 : 800 = 2,28125%

a) 

2Al + 6HCl -> 2AlCl3 + 3H2

  2           6             2            3

0,19       0,57        0,19       0,285

n_Al= m/M = 5,003/27= 0,19 mol

b) 

V_H2= n . 22,4 = 0,285 . 22,4 = 6,384 lít

c) C_{M HCl}= n/V = 0,57 / 0,2 = 2,85M

d)     H₂      +     CuO    ->    Cu    +    H₂O ( điều kiện nhiệt độ nha mik k đánh kí hiệu lên mũi tên đc )

         1                  1              1               1

                                         0,15625

n_CuO= 12,5 / 80 = 0,15625 mol

=> tính số mol theo CuO

m_Cu= n. M = 0,15625 . 64 = 10g

bạn để bài mấy được k

Zn + 2HCl --> ZnCl2 + H2

0,5      1              0,5       0,5

nZn=\(\dfrac{32,5}{65}\)=0,5(mol)

mZnCl2=0,5.136=68(g)

VH2=0,5.22,4=11,2(l)

c) CM HCl=\(\dfrac{1}{0,25}=4M\)